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O “K” : Manipulation of K Kp verse Kc.  Write an equilibrium constant expression for any chemical reaction. The concentrations of solids and solvents.

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Presentation on theme: "O “K” : Manipulation of K Kp verse Kc.  Write an equilibrium constant expression for any chemical reaction. The concentrations of solids and solvents."— Presentation transcript:

1 o “K” : Manipulation of K Kp verse Kc

2  Write an equilibrium constant expression for any chemical reaction. The concentrations of solids and solvents (e.g., water) are not included in equilibrium constant expressions  Recognize that a large value of K (K>> 1) means the reaction is product-favored. A small value of K (K << 1) indicates a reactant-favored reaction.  Use equilibrium constants to calculate the concentration of a reactant or product at equilibrium.  Apply Le Chatelier’s principle to predict the effect of a disturbance on a chemical equilibrium: a change in temperature, a change in concentrations, or a change in volume or pressure for a reaction involving gases

3 Calculate an equilibrium constant given the reactant and product concentrations at equilibrium Apply the idea of the reaction quotient (Q) to decide if a reaction is at equilibrium. Q = K; reaction is at equilibrium Q K; products convert to reactants to attain equilibrium

4  K is designated as K c when Equilibrium concentrations are expressed in terms of reactant and product concentration, M (moles per liter).  K is designated as K p, when concentrations of gases are represented by partial pressures (amount of gas measured in atm or Kpa)

5  Know how K changes as different stoichiometric coefficients are used in a balanced equation and when equation is reversed  Know that, when two chemical equations are added to give a net equation, the value of K for the net equation is the product of the values of K for the summed equations

6 Kp

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8 2NO 2 + O 2 ↔ 2NO 3 + C  +4 2H 2 O + C H 4  CO 2 +4 H 2 H 2 + I 2 ↔ 2HI

9 S(s) + O 2 (g) ---> SO 2 (g) Fe 3+ (aq) + SCN - (aq)  FeSCN N  2N 3 H 2 (g) + N 2 (g)  2N H 3 (g)

10 Kp verse Kc

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14 K 3 = K 1 * K 2 K 3 = K 1 K 2 K w = K a * K b

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